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2 years ago

Give me lil reasoning so I know your not lying for points

Chemistry

1 answer:

ycow [4]2 years ago

4 0

Answer:

0.87 ATM

Explanation:

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Boyle's law states that the pressure of a gas is always:

natka813 [3]

Answer:

d. inversely proportional to the volume of its container.

Explanation:

Boyle's law states that at constant temperature and number of moles, the pressure of the gas is inversely proportional to the volume of the gas.

Thus, P ∝ T

P is the pressure

T is the temperature

For two gases at same temperature, the law can be written as:-

${P_1}\times {V_1}={P_2}\times {V_2}$

<u>Thus, according to the question, the answer is:- d. inversely proportional to the volume of its container.</u>

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2 years ago

Will essential oil penetrate through clothes

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Https://wellnessmama.com/26519/risks-essential-oils/ is what u can learn maybe u can find ue answer Hope i helped best of luck :)

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2 years ago

Read 2 more answers

Place the following in order of increasing X-Se-X bond angle, where X represents the outer atoms in each molecule. SeO2 SeCl6 Se

GalinKa [24]

Answer:

SeCl₆ < SeF₂ < SeO₂

Explanation:

(A) SeO₂

The central atom has 2 bond pairs and 1 lone pair. The molecule is bent shaped which has an angle of 120°.

(B) SeCl₆

The central atom has 6 bond pairs and 0 lone pair. The geometry is octahedral in which the equatorial bonds has an angle of 90° and axial bond has an angle of 90°.

The central atom has 2 bond pairs and 2 lone pairs. The geometry is bent shape which has an angle of approximately 105.5°.

The order is:

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3 years ago

Coal containing 15.0% H2O, 2.0% S and 83.0% C by mass is burnt with the stoichiometric amount of air in a furnace. What is the m

devlian [24]

Answer:

This is a coal combustion process and we will assume

Inlet coal amount = 100kg

It means that there are

15kg of H2O, 2kg of Sulphur and 83kg of Carbon

Now to find the mole fraction of SO2(g) in the exhaust?

Molar mass of S = 32kg/kmol

Initial moles n of S = 2/32 = 0.0625kmols

Reaction: S + O₂ = SO₂

That is 1 mole of S reacts with 1 mole of O₂ to give 1 mole of SO₂

Then, it means for 0.0625 kmoles of S, we will have 0.0625 kmole of SO2 coming out of the exhaust

The mole fraction of SO2(g) in the exhaust=0.0625kmols

Explanation:

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2 years ago

50.0 mL solution of 0.160 M potassium alaninate ( H 2 NC 2 H 5 CO 2 K ) is titrated with 0.160 M HCl . The p K a values for the

scZoUnD [109]

Answer:

a) 6.12

b) 1.87

Explanation:

At the onset of the equivalence point (i.e the first equivalence point); alaninate is being converted to alanine.

$H_2NC_2H_5CO^-_2$ $+$ $H^+$ ------> $H_3}^+NC_2H_5CO^-_2$

1 mole of alaninate react with 1 mole of acid to give 1 mole of alanine;

therefore 50.0 mL of 0.160 M alaninate required 50.0 mL of 0.160M HCl to reach the first equivalence point.

The concentration of alanine can be gotten via the following process as shown below;

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{initial moles of alaninate}{total volume}$

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{(50.0mL)*(0.160M)}{(50.0mL+50.0mL)}$

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{8}{100mL}$

$[H_3}^+NC_2H_5CO^-_2]$ = 0.08 M

Alanine serves as an intermediary form, however the concentration of $H^+$ and the pH can be determined as follows;

$[H^+]$ = $\sqrt{\frac{K_{a1}K_{a2}{[H_3}^+NC_2H_5CO^-_2]+K_{a1}K_w}{ K_{a1}{[H_3}^+NC_2H_5CO^-_2] } }$

$[H^+]$ = $\sqrt{\frac{ (10^{-pK_{a1})}(10^{-pK_{a2})}(0.08)+(10^{-pK_{a1})}(1.0*10^{-14})} {(10^{-pK_{a1}})+(0.08)} }$

$[H^+]$ = $\sqrt{\frac{ (10^{-2.344})(10^{-9.868})(0.08)+(10^{-2.344})(1.0*10^{-14})} {(10^{-2.344})+(0.08)} }$

$[H^+]$ = $7.63*10^{-7}M$

pH = - log $[H^+]$

pH = $-log[7.63*10^{-7}]$

pH= 6.12

Therefore, the pH of the first equivalent point = 6.12

b) At the second equivalence point; all alaninate is converted into protonated alanine.

$H_2NC_2H_5CO^-_2$ $+$ $H^+$ -----> $H^+_3NC_2H_5CO^-_2$

$H^+_3NC_2H_5CO^-_2$ $+$ $H^+$ -----> $H^+_3NC_2H_5CO_2H$

Here; we have a situation where 1 mole of alaninate react with 2 moles of acid to give 1 mole of protonated alanine;

Moreover, 50.0 mL of 0.160 M alaninate is needed to produce 100.0mL of 0.160 M HCl in order to achieve the second equivalence point.

Thus, the concentration of protonated alanine can be determined as:

$[H^+_3NC_2H_5CO_2H]$ = $\frac{initial moles of alaninate}{total volume}$

$[H^+_3NC_2H_5CO_2H]$ = $\frac{(50.0mL)*(0.160M)}{(50.0mL+100.0mL)}$

$[H^+_3NC_2H_5CO_2H]$ = $\frac{8}{150}$

$[H^+_3NC_2H_5CO_2H]$ = 0.053 M

The pH at the second equivalence point can be calculated via the dissociation of protonated alanine at equilibrium which is represented as:

$H^+_3NC_2H_5CO_2H$ ⇄ $H^+_3NC_2H_5CO^-_2$ $+$ $H^+$

(0.053 - x) x x

$K_{a1}$ = $\frac{[H^+] [H^+_3NC_2H_5CO^-_2]}{[H^+_3NC_2H_5CO_2H]}$

$10^{-PK_{a1}}$ = $\frac{x*x}{(0.053-x)}$

$10^{-2.344} =\frac{x^2}{(0.053-x)}$

$0.00453 = \frac{x^2}{(0.053-x)}$

$0.00453(0.053-x) =x^2$

$x^2+0.00453x-(2.4009*10^{-4})$

Using quadratic equation formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

we have:

$\frac{-0.00453+\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}$ OR $\frac{-0.00453-\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}$

= 0.0134 OR -0.0179

So; we go by the positive integer which says

x = 0.0134

So $[H^+]=[H_3^+NC_2H_5CO^-_2]= 0.0134 M$

pH = $-log[H^+]$

pH = $-log[0.0134]$

pH = 1.87

Thus, the pH of the second equivalent point = 1.87

3 0

3 years ago